I would like to prove that:

If $m,n \in \mathbb{N}$ then $\sqrt[m]{n}$ is either an integer or irrational.

Here is my attempt:

Suppose that $\sqrt[m]{n} \in \mathbb{Q}$. Then $\exists \; p,q \in \mathbb{Z}$ such that

$$ \sqrt[m]{n} = \frac{p}{q}$$

which implies that $$ n = \frac{p^m}{q^m}.$$

Now if $q \neq 1$ then $q^m | p^m$ since $n \in \mathbb{N}$.

Because $\sqrt[m]{n} \in \mathbb{Q}$ then we can assume that $gcd(p,q)=1.$

By the fundamental theorem of arithmetic we have that $gcd(p,q)=gcd(p^m,q^m)$ hence

$$ gcd(p^m,q^m)=1 \implies q^m \nmid p^m \implies q =1$$

so $n=p^m \implies \sqrt[m]{n} =p \in \mathbb{Z}$ which contradicts the fact that $\sqrt[m]{n} \in \mathbb{Q}$.

Therefore $\forall \; m,n \in \mathbb{N}$ if $\sqrt[m]{n}$ is not an integer then it is not rational.